Mutually Independent Events

Why

We can extend the notion of independnence beyond pairs of uncertain events, to sets of events.

Definition

Suppose $P$ is a event probability function on a finite sample space $\Omega $. The events $A_1, \dots , A_n$ are independent (or mutually independent), if for all $k$ between 1 and $n$, and distinct indicies $i_1, \dots , i_k$ between $1$ and $n$,

\[ P(A_{i_1} \cap A_{i_2} \cap \cdots \cap A_{i_k}) = P(A_{i_1}) P(A_{i_2}) \cdots P(A_{i_k}) . \]

Similar to the case of pairs of events, one can show that this condition is equivalent to the statement that for any distinct indices $i_1, \dots , i_k, j_1, \dots , j_l$,

\[ P(A_{j_1} \cap \cdots \cap A_{j_l} \mid A_{i_1} \cap \cdots \cap A_{i_k}) = P(A_{j_1} \cap \cdots \cap A_{j_l}) \]

Examples

$n$ tosses of a coin. As usual, model $n$ tosses of a coin with $\set{0,1}^n$ and put a distribution $p: \Omega \to [0,1]$ so that

\[ p(\omega ) = 1/2^n \quad \text{for all } \omega \in \Omega \]

Now, for $i = 1, \dots , n$, define the event $A_i$ by

\[ A_i = \Set{\omega \in \Omega }{\omega (i) = 1} \]

We claim that the set $\set{A_1,. \dots , A_n}$ is mutually independent. To see this, notice that for any distinct indices $i_1, \dots , i_k$,

\[ \num{A_{i_1} \cap \cdots \cap A_{i_k}} = 2^{n-k} \]

This holds because, once $k$ of the coin flips, there are $2^{n-k}$ ways for the remaining coins to land (using the fundamental principle of counting). Consequently,

\[ P(A_{i_1} \cap \cdots \cap A_{i_k}) = \frac{2^{n-k}}{2^n} = 2^{-k} \]

We can use this result with one set $P(A_i) = 1/2$, and so we obtain

\[ P(A_{i_1} \cap \cdots A_{i_k}) = P(A_{i_1}) \cdots P(A_{i_k}), \]

as desired.

Basic implications

It can be shown¹ that if $A_1, \dots , A_n$ are indepnednet events, and $B_1, \dots , B_n$ are events such that $B_i$ is either $A_i$ or $A_i^c$, then $B_1, \dots , B_n$ are mutually independent.

Future editions will. ↩︎